Інтеграл від $x^{3} - 3 x^{2}$

Знайдіть $\int \left(x^{3} - 3 x^{2}\right)\, dx$.

Integrate term by term:

$${\color{red}{\int{\left(x^{3} - 3 x^{2}\right)d x}}} = {\color{red}{\left(- \int{3 x^{2} d x} + \int{x^{3} d x}\right)}}$$

Apply the power rule $\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$ $\left(n \neq -1 \right)$ with $n=3$:

$$- \int{3 x^{2} d x} + {\color{red}{\int{x^{3} d x}}}=- \int{3 x^{2} d x} + {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- \int{3 x^{2} d x} + {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=3$ and $f{\left(x \right)} = x^{2}$:

$$\frac{x^{4}}{4} - {\color{red}{\int{3 x^{2} d x}}} = \frac{x^{4}}{4} - {\color{red}{\left(3 \int{x^{2} d x}\right)}}$$

Apply the power rule $\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$ $\left(n \neq -1 \right)$ with $n=2$:

$$\frac{x^{4}}{4} - 3 {\color{red}{\int{x^{2} d x}}}=\frac{x^{4}}{4} - 3 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\frac{x^{4}}{4} - 3 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Тому,

$$\int{\left(x^{3} - 3 x^{2}\right)d x} = \frac{x^{4}}{4} - x^{3}$$

Спростити:

$$\int{\left(x^{3} - 3 x^{2}\right)d x} = \frac{x^{3} \left(x - 4\right)}{4}$$

Додайте константу інтегрування:

$$\int{\left(x^{3} - 3 x^{2}\right)d x} = \frac{x^{3} \left(x - 4\right)}{4}+C$$

Answer: $\int{\left(x^{3} - 3 x^{2}\right)d x}=\frac{x^{3} \left(x - 4\right)}{4}+C$